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Question
Answer the following :
Find the equation of tangent to the circle x2 + y2 = 64 at the point `"P"((2pi)/3)`
Solution
Given equation of circle is
x2 + y2 = 64
Comparing this equation with x2 + y2 = r2, we get
r = 8
The equation of a tangent to the circle
x2 + y2 = r2 at P(θ) is
x cosθ + y sinθ = r
∴ the equation of the tangent at `"P"((2pi)/3)` is
`xcos (2pi)/3 + y sin (2pi)/3` = 8
∴ `x((-1)/2) + y(sqrt(3)/2)` = 8
∴ `-x + sqrt(3)y` = 16
∴ `x - sqrt(3)y + 16` = 0.
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